Found problems: 3585
Circles $\omega_1$ and $\omega_2$ intersect each other at points $A$ and $B$. Point $C$ lies on the tangent line from $A$ to $\omega_1$ such that
$\angle ABC = 90^\circ$. Arbitrary line $\ell$ passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut $\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot of altitude from $A$ to $\ell$. Prove that points $X, Y$ and $Z$ are collinear.
[i]Proposed by Iman Maghsoudi[/i]
The altitudes $AH_1,BH_2,CH_3$ of an acute-angled triangle $ABC$ meet at point $H$. Points $P$ and $Q$ are the reflections of $H_2$ and $H_3$ with respect to $H$. The circumcircle of triangle $PH_1Q$ meets for the second time $BH_2$ and $CH_3$ at points $R$ and $S$. Prove that $RS$ is a medial line of triangle $ABC$.
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
In convex quadrilateral $ABCD$, let $M$ and $N$ denote the midpoints of sides $AD$ and $BC$, respectively. On sides $AB$ and $CD$ are points $K$ and $L$, respectively, such that $\angle MKA=\angle NLC$. Prove that if lines $BD$, $KM$, and $LN$ are concurrent, then \[ \angle KMN = \angle BDC\qquad\text{and}\qquad\angle LNM=\angle ABD.\]
$ABCD$ is a convex quadrilateral where $\angle A = 45^\circ$ and $\angle C = 135^\circ$. $P$ is a point strictly inside $\triangle ABC$ such that $\angle BAP = \angle CAD$ and $\angle BCP = \angle ACD$. Prove that $PB \perp PD$ if and only if $AC \perp BD$.
Given an acute triangle $ABC$ with circumcenter $O$. The circumcircle of $BCH$ and a circle with diameter of $AC$ intersect at $P (P \neq C)$. A point $Q$ on segment of $PC$ such that $PB = PQ$. Prove that $\angle ABC = \angle AQP$
Let $ABCD$ be a cyclic quadrilateral, $K$ and $N$ be the midpoints of the diagonals and $P$ and $Q$ be points of intersection of the extensions of the opposite sides. Prove that $\angle PKQ + \angle PNQ = 180$.
[i]($7$ points)[/i]
.
Let ABCD is a cyclic quadrilateral inscribed in $(O)$. $E, F$ are the midpoints of arcs $AB$ and $CD$ not containing the other vertices of the quadrilateral. The line passing through $E, F$ and parallel to the diagonals of $ABCD$ meet at $E, F, K, L$. Prove that $KL$ passes through $O$.
Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
A convex hexagon $ABCDEF$ with an interior point $P$ is given. Assume that $BCEF$ is a square and both $ABP$ and $PCD$ are right isosceles triangles with right angles at $B$ and $C$, respectively. Lines $AF$ and $DE$ intersect at $G$. Prove that $GP$ is perpendicular to $BC$.
[i]Proposed by Patrik Bak - Slovakia[/i]
In triangle $ABC$ $\angle B = 60^o, O$ is the circumcenter, and $L$ is the foot of an angle bisector of angle $B$. The circumcirle of triangle $BOL$ meets the circumcircle of $ABC$ at point $D \ne B$. Prove that $BD \perp AC$.
(D. Shvetsov)
$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear.
[i]V. Astakhov[/i]
Let $M$ be the intersection of the diagonals $AC$ and $BD$ of cyclic quadrilateral $ABCD$. If $|AB|=5$, $|CD|=3$, and $m(\widehat{AMB}) = 60^\circ$, what is the circumradius of the quadrilateral?
$
\textbf{(A)}\ 5\sqrt 3
\qquad\textbf{(B)}\ \dfrac {7\sqrt 3}{3}
\qquad\textbf{(C)}\ 6
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ \sqrt{34}
$
Let $ABC$ be an acute non-isosceles triangle with circumcircle $\omega$, circumcenter $O$ and orthocenter $H$. We draw a line perpendicular to $AH$ through $O$ and a line perpendicular to $AO$ through $H$. Prove that the points of intersection of these lines with sides $AB$ and $AC$ lie on a circle, which is tangent to $\omega$.
[i]Proposed by A. Kuznetsov[/i]
Let a triangle $ABC$ , $O$ it's circumcenter , $H$ ortocenter and $M$ the midpoint of $AH$. The perpendicular at $M$ to line $OM$ meets $AB$ and $AC$ at points $P$, respective $Q$. Prove that $MP=MQ$.
Babis
Let $ABC$ a triangle with circumcircle $\Gamma$ and circumcenter $O$. A point $X$, different from $A$, $B$, $C$, or their diametrically opposite points, on $\Gamma$, is chosen. Let $\omega$ the circumcircle of $COX$. Let $E$ the second intersection of $XA$ with $\omega$, $F$ the second intersection of $XB$ with $\omega$ and $D$ a point on line $AB$ such that $CD \perp EF$.
Prove that $E$ is the circumcenter of $ADC$ and $F$ is the circumcenter of $BDC$.
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.
Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.
[i]Proposed by Anton Trygub, Ukraine[/i]
Rhombus $PQRS$ is inscribed in rectangle $ABCD$ so that vertices $P$, $Q$, $R$, and $S$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB=15$, $BQ=20$, $PR=30$, and $QS=40$. Let $m/n$, in lowest terms, denote the perimeter of $ABCD$. Find $m+n$.
Let $\gamma$ be the incircle of $\triangle ABC$ (i.e. the circle inscribed in $\triangle ABC$) and $I$ be the center of $\gamma$. Let $D$, $E$ and $F$ be the feet of the perpendiculars from $I$ to $BC$, $CA$, and $AB$ respectively. Let $D'$ be the point on $\gamma$ such that $DD'$ is a diameter of $\gamma$. Suppose the tangent to $\gamma$ through $D$ intersects the line $EF$ at $P$. Suppose the tangent to $\gamma$ through $D'$ intersects the line $EF$ at $Q$. Prove that $\angle PIQ + \angle DAD' = 180^{\circ}$.
Let $ABC$ be an acute triangle and $D, E, F$ are the midpoints of $BC, CA, AB$, respectively. The circle with diameter $AD$ intersects the lines $AB$ and $AC$ at points $P$ and $Q$ , respectively. The lines through $P$ and $Q$ parallel to $BC$ intersect $DE$ at point $R$ and $DF$ at point $S$, respectively. The circumcircle of $DPR$ intersects $AB$ at $X$, the circumcircle of $DQS$ intersects $AC$ in $Y$, and these two circles intersect again point $Z$. Prove that $Z$ is the midpoint of $XY$.
Given is an acute triangle $ABC$. Its heights $BB_1$ and $CC_1$ are extended past points $B_1$ and $C_1$. On these extensions, points $P$ and $Q$ are chosen, such that angle $PAQ$ is right. Let $AF$ be a height of triangle $APQ$. Prove that angle $BFC$ is a right angle.
The circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent line of the two circles closest to point $D$ touches $\Gamma_1$ in A and $\Gamma_2$ in $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the midpoint of line segment $BC$. Prove that $\angle DPM = \angle BDC$.
Let the circles $\omega$ and $\omega^ \prime$ intersect in $A$ and $B$. Tangent to circle$\omega$ at $A$ intersects$\omega^ \prime$ in $C$ and tangent to circle $\omega^ \prime$ at $A$ intersects $\omega$ in $D$. Suppose that $CD$ intersects$\omega$ and $\omega^ \prime$ in $E$ and $F$, respectively (assume that $E$ is between $F$ and $C$). The perpendicular to $AC$ from $E$ intersects $\omega^ \prime$ in point $P$ and perpendicular to $AD$ from $F$ intersects$\omega$ in point $Q$ (The points $A, P$ and $Q$ lie on the same side of the line $CD$). Prove that the points $A, P$ and $Q$ are collinear.
Proposed by Mahdi Etesami Fard
Let $ABC$ a right isosceles triangle with hypotenuse $AB=\sqrt2$ . Determine the positions of the points $X,Y,Z$ on the sides $BC,CA,AB$ respectively so that the triangle $XYZ$ is isosceles, right, and with minimum area.